How to find exact solution of this volterra equation?

Question

I was working on numerical solution of this equation (by block pulse). $$x(t)=1+\int_{0}^{t}s^2x(s)ds+\int_{0}^{t}sx(s)dB(s)\\t \in[0,\frac{1}{2}]$$B(t) is standard brownian motion. Author of the paper said :the exact solution of this equation is $$ x(t)=e^{\frac{t^3}{6}+\int_{0}^{t}sdB(s)}$$ I tried to check the exact solution into it ,.I tried to find the exact solution ,but I can't.please help me to understand how obtain this exact solution (or check into equation).

Answer 1

Let $Y_t = \int_0^ts\,dB_s$. Note that $Y$ is a martingale and hence a semimartingale. Then consider the process $X_t = f(t,Y_t)$ with $f(t,x) = e^{\frac{t^3}{6}+x}$. Since the partial derivatives $f_t,f_x,f_{xx}$ all exist and are continuous, Ito's lemma applies in this situation. $$f(t,Y_t) = f(0,Y_0) + \int_0^tf_t(s,Y_s)\,ds + \int_0^tf_x(s,Y_s)\,dY_s + \frac{1}{2}\int_0^tf_{xx}(s,Y_s)\,d[Y,Y]_s$$ It is easy to compute that $[Y,Y]_t = \int_0^ts^2\,ds = \frac{t^3}{3}$. Furthermore, $dY_s = s\,dB_s$ and $X_0 = f(0,Y_0) = 1$. Putting everything together $$X_t = X_0 + \frac{1}{2}\int_0^ts^2X_s\,ds + \int_0^tsX_s\,dB_s + \frac{1}{2}\int_0^ts^2X_s\,ds = 1 + \int_0^tsX_s\,dB_s + \int_0^ts^2X_s\,ds$$

To show $[Y,Y]_t = \int_0^ts^2\,ds$, I used the following result.

For semimartingales $M,N$ $$[\int G_sdM_s,\int H_sdN_s]_t = \int_{(0,t]}G_sH_s\,d[M,N]_s$$